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Draw and explain the phasor diagram of $1-\phi$
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  • When the transformer is operating at no load there is iron loss in the core and the copper loss in primary winding.
  • Thus primary input current I0 has to supply iron loss in the core and a very small amount of copper loss in priamry. Hence the current I0 has two components;
  1. a magnitising or reactive component$I_{\mu }\ and $
  2. power or active component Iw
  • The magnetising component$I_{\mu }$ is responsible for setting up fluxes in the core. It is in phase with the flux$\phi $.

              $I_{\mu }=I_o\ sin\ \phi _o$

  • The active component Iw is responsible for power loss in the transformer. It is in phase with V1.

               $I_w=I_o\ cos\phi _o$

  • Hence, no load current Io is the phaser sum of$I_{\mu }\ and \ I_w. $

               $I_o=I_{\mu }+I_w\\ I_o=\sqrt {I^2_{\mu }+I^2 w}$

  • The no load current Io is very small as compared to full load current I1. Hence copper loass is negligible and no load input power is practically equal to iron loss or core loss in the transformer.

  • Iron loss W1 = V1 Io cos$\phi _o$ where,$cos\ \phi_o\ $ is power factor at no load.

Phasor Diagram: The phasor diagram is shown in Fig. 1.

Fig. 1 Phasor Diagram

  • Since the flux$\phi $ is common to both the windings,$\phi $ is chosen as reference phasor.
  • From emf equation of the transformer, it is clear that E1 and E2 lag the flux by 900. Hence emfs E1 and E2 are drawn such that these lag behind the flux$\phi $ by 90o.
  • The magnitising component$I_{\mu }$ is drawn in phase with the flux$\phi. $
  • The applied Voltage is drawn equal and opposite to E1 as$V_1\approx E_1.$
  • The active component Iw is drawn in phase with the voltage V1.
  • The phasor sum of$I_{\mu }\ and\ I_w\ $ gives the no load current Io.
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